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How can I decompose the following function into partial fractions in order to integrate it? $$\frac{1}{v(\ln(v) - 2)}$$

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Partial fraction decomposition is appropriate to use for rational functions $f(x)$, i.e., when there exists a polynomial $p(x)$ and a polynomial $q(x)$ where the degree of $q(s) > $ the degree of $p(x)$, and such that $$f(x) = \frac{p(x)}{q(x)}.$$

But the function you posted is not a rational function, precisely because the denominator $\;v\ln(v)-2\;$ is not a polynomial.

Consider, instead using substitution to integrate: $$\text{Let }\,\big(\ln(v) -2\big) = x.$$

Then $$\frac{1}{v}\,dv = dx$$

That gives us $$\int\frac{\,dv}{v(\ln(v) - 2)} = \int \frac {dx}{x} =\; \ln (x) + C \;= \;\ln \big(\ln (v)-2\big)+C$$

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